## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

## Holooly Tables

All the data tables that you may search for.

## Holooly Help Desk

Need Help? We got you covered.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 7.5

A rectangular beam 300 mm deep is simply supported over a span of 4 metres. Determine the uniformly distributed load per metre which the beam may carry, if the bending stress should not exceed 120 N/mm². Take $I = 8 × 10^6 mm^4.$

## Verified Solution

Given :
Depth,       d = 300 mm
Span,         L = 4 m
Max. bending stress, $σ_{\max} = 120 N/mm^2$
Moment of inertia,    $I = 8 × 10^6 mm^4$
Let, w = U.D.L. per metre length over the beam in N/m.

The bending stress will be maximum, where bending moment is maximum. For a simply supported beam carrying U.D.L., the bending moment is maximum at the centre of the beam [i.e., at point C of Fig. 7.11 (b)]

∴    Max. B.M. = 2w × 2 – 2w × 1 = 4w – 2w
= 2w Nm $\left(\text { Also } M=\frac{w \times L^2}{8}=\frac{w \times 4^2}{8}=\frac{16 w}{8}=2 w\right)$
= 2w × 1000 Nmm
or           M = 2000w Nmm

Now using equation (7.6), we get

$M=\sigma_{\max } \times Z$            …(i)

where          $Z=\frac{I}{y_{\max }}=\frac{8 \times 10^6}{150}\quad \left(\because y_{\max }=\frac{d}{2}=\frac{300}{2}=150 mm \right)$

Hence above equation (i) becomes as

$2000 w=120 \times \frac{8 \times 10^6}{150}$

or            $w=\frac{120 \times 8 \times 10^6}{2000 \times 150}=\pmb{3200 N / m.}$ 